How to find the general solution of the equation sin θ = 0?

Prove that the general solution of sin θ = 0 is θ = nπ, n ∈ Z

**Solution:**

According to the figure, by definition, we have,

Sine function is defined as the ratio of the side opposite divided by the hypotenuse.

Let O be the centre of a unit circle. We know that in unit circle, the length of the circumference is 2π.If we started from A and moves in anticlockwise direction then at the points A, B, A', B' and A, the arc length travelled are 0, \(\frac{π}{2}\), π, \(\frac{3π}{2}\), and 2π.

Therefore, from the above unit circle it is clear that

sin θ = \(\frac{PM}{OP}\)

Now, sin θ = 0

⇒ \(\frac{PM}{OP}\) = 0

⇒ PM = 0.

So when will the sine be equal to zero?

Clearly, if PM = 0 then the final arm OP of the angle θ coincides with OX or, OX'.

Similarly, the final arm OP coincides with OX or OX' when θ = 0, π, 2π, 3π, 4π, 5π …………….., -π, , -2π, -3π, -4π, -5π ………., i.e., when θ = 0 or an integral multiples of π i.e., when θ = nπ where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Hence, **θ = nπ**, n ∈
Z is the general solution of the given equation sin θ = 0

**1.** Find the general solution of the equation sin 2θ = 0

**Solution: **

sin 2θ = 0

⇒ 2θ = nπ, where, n = 0, ± 1, ± 2, ± 3,……., [Since, we know that **θ = nπ**, n ∈
Z is the general solution of the given equation sin θ = 0]

⇒ θ = \(\frac{nπ}{2}\), where, n = 0, ± 1, ± 2, ± 3,…….

Therefore, the general solution of the equation sin 2θ = 0 is θ = \(\frac{nπ}{2}\), where, n = 0, ± 1, ± 2, ± 3,…….

**2.** Find the general solution of the equation sin \(\frac{3x}{2}\) = 0

**Solution: **

sin \(\frac{3x}{2}\) = 0

⇒ \(\frac{3x}{2}\) = nπ, where, n = 0, ± 1, ± 2, ± 3,…….[Since, we know that **θ = nπ**, n ∈
Z is the general solution of the given equation sin θ = 0]

⇒ x = \(\frac{2nπ}{3}\), where, n = 0, ± 1, ± 2, ± 3,…….

Therefore, the general solution of the equation sin \(\frac{3x}{2}\) = 0 is θ = \(\frac{2nπ}{3}\), where, n = 0, ± 1, ± 2, ± 3,…….

3. Find the general solution of the equation tan 3x = tan 2x + tan x

**Solution: **

tan 3x = tan 2x + tan x

⇒ \(\frac{sin 3x}{cos 3x}\) = \(\frac{sin 2x}{cos 2x}\) + \(\frac{sin x}{cos x}\)

⇒ \(\frac{sin 3x}{cos 3x}\) = \(\frac{sin 2x cos x + cos 2x sin x}{cos 2x cos x}\)

⇒ cos 3θ sin (2x + x) = sin 3x cos 2x cos x

⇒ cos 3x sin 3x = sin 3x cos 2x cosx

⇒ cos 3x sin 3x - sin 3x cos
2x cos x = 0

⇒ sin 3x [cos (2x + x) - cos 2x cos x] = 0

⇒ sin 3x . sin 2x sin x = 0

Either either, sin 3x = 0 or, sin 2x = 0 or, sin x = 0

⇒ 3x = nπ or, 2x = nπ or, x = nπ

⇒ x = \(\frac{nπ}{3}\) …..... (1) or, x = \(\frac{nπ}{2}\) …..... (2) or, x = nπ …..... (3), where n ∈ I

Clearly, the value of x given in (2) are∶ 0, \(\frac{π}{2}\), π, \(\frac{3π}{2}\), 2π, \(\frac{5π}{2}\) ……………., - \(\frac{π}{2}\),- π, - \(\frac{3π}{2}\) , …………

It is readily seen that the solution x = \(\frac{π}{2}\), \(\frac{3π}{2}\), \(\frac{5π}{2}\)………, - \(\frac{π}{2}\), - \(\frac{3π}{2}\),………

Of the above solution do not satisfy the given equation.

Further to not that the rest solutions of (2) and the solution of (3) are contained in the solutions (1).

Therefore, the general solution of the equation tan 3x = tan 2x + tan x is x = \(\frac{3π}{2}\),, where n ∈ I

**4.** Find the general solution of the equation sin\(^{2}\) 2x = 0

**Solution: **

sin\(^{2}\) 2x = 0

sin 2x = 0

⇒ 2x = nπ, where, n = 0, ± 1, ± 2, ± 3,……., [Since, we know that **θ = nπ**, n ∈
Z is the general solution of the given equation sin θ = 0]

⇒ x = \(\frac{nπ}{2}\), where, n = 0, ± 1, ± 2, ± 3,…….

Therefore, the general solution of the equation sin\(^{2}\) 2x = 0 is x = \(\frac{nπ}{2}\), where, n = 0, ± 1, ± 2, ± 3,…….

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